The Stacks project

Proof. Let $\mathcal{D} \subset K(A, \text{d})$ be the triangulated subcategory discussed in Remark 22.36.1. Let $P$ be an object of $\mathcal{D}$ which is finite projective as a graded $A$-module. Then $P$ represents a compact object of $D(A, \text{d})$ by Remark 22.36.2.

To prove the converse, let $E$ be a compact object of $D(A, \text{d})$. Fix $a \leq b$ as in Lemma 22.36.5. After decreasing $a$ and increasing $b$ if necessary, we may also assume that $H^ i(E) = 0$ for $i \not\in [a, b]$ (this follows from Proposition 22.36.4 and our assumption on $A$). Moreover, fix an integer $c > 0$ such that $A^ n = 0$ if $|n| \geq c$.

By Proposition 22.36.4 we see that $E$ is a direct summand, in $D(A, \text{d})$, of a differential graded $A$-module $P$ which has a finite filtration $F_\bullet$ by differential graded submodules such that $F_ iP/F_{i - 1}P$ are finite direct sums of shifts of $A$. In particular, $P$ has property (P) and we have $\mathop{\mathrm{Hom}}\nolimits _{D(A, \text{d})}(P, M) = \mathop{\mathrm{Hom}}\nolimits _{K(A, \text{d})}(P, M)$ for any differential graded module $M$ by Lemma 22.22.3. In other words, $P$ is an object of the triangulated subcategory $\mathcal{D} \subset K(A, \text{d})$ discussed in Remark 22.36.1. Note that $P$ is finite free as a graded $A$-module.

Choose $n > 0$ such that $b + 4c - n < a$. Represent the projector onto $E$ by an endomorphism $\varphi : P \to P$ of differential graded $A$-modules. Consider the distinguished triangle

in $K(A, \text{d})$ where $C$ is the cone of the first arrow. Then $C$ is an object of $\mathcal{D}$, we have $C \cong E \oplus E[1]$ in $D(A, \text{d})$, and $C$ is a finite graded free $A$-module. Next, consider a distinguished triangle

in $K(A, \text{d})$ where $C'$ is the cone on a morphism $C[1] \to C$ representing the composition

in $D(A, \text{d})$. Then we see that $C'$ represents $E \oplus E[2]$. Continuing in this manner we see that we can find a differential graded $A$-module $P$ which is an object of $\mathcal{D}$, is a finite free as a graded $A$-module, and represents $E \oplus E[n]$.

Choose a basis $x_ i$, $i \in I$ of homogeneous elements for $P$ as an $A$-module. Let $d_ i = \deg (x_ i)$. Let $P_1$ be the $A$-submodule of $P$ generated by $x_ i$ and $\text{d}(x_ i)$ for $d_ i \leq a - c - 1$. Let $P_2$ be the $A$-submodule of $P$ generated by $x_ i$ and $\text{d}(x_ i)$ for $d_ i \geq b - n + c$. We observe

1. $P_1$ and $P_2$ are differential graded submodules of $P$,

2. $P_1^ t = 0$ for $t \geq a$,

3. $P_1^ t = P^ t$ for $t \leq a - 2c$,

4. $P_2^ t = 0$ for $t \leq b - n$,

5. $P_2^ t = P^ t$ for $t \geq b - n + 2c$.

As $b - n + 2c \geq a - 2c$ by our choice of $n$ we obtain a short exact sequence of differential graded $A$-modules

Since $P$ is projective as a graded $A$-module this is an admissible short exact sequence (Lemma 22.16.1). Hence we obtain a boundary map $\delta : P \to (P_1 \cap P_2)[1]$ in $K(A, \text{d})$, see Lemma 22.7.2. Since $P = E \oplus E[n]$ and since $P_1 \cap P_2$ lives in degrees $(b - n, a)$ we find that $\mathop{\mathrm{Hom}}\nolimits _{D(A, \text{d})}(E \oplus E[n], (P_1 \cap P_2)[1])$ is zero. Therefore $\delta = 0$ as a morphism in $K(A, \text{d})$ as $P$ is an object of $\mathcal{D}$. By Derived Categories, Lemma 13.4.11 we can find a map $s : P \to P_1 \oplus P_2$ such that $\pi \circ s = \text{id}_ P + \text{d}h + h\text{d}$ for some $h : P \to P$ of degree $-1$. Since $P_1 \oplus P_2 \to P$ is surjective and since $P$ is projective as a graded $A$-module we can choose a homogeneous lift $\tilde h : P \to P_1 \oplus P_2$ of $h$. Then we change $s$ into $s + \text{d} \tilde h + \tilde h \text{d}$ to get $\pi \circ s = \text{id}_ P$. This means we obtain a direct sum decomposition $P = s^{-1}(P_1) \oplus s^{-1}(P_2)$. Since $s^{-1}(P_2)$ is equal to $P$ in degrees $\geq b - n + 2c$ we see that $s^{-1}(P_2) \to P \to E$ is a quasi-isomorphism, i.e., an isomorphism in $D(A, \text{d})$. This finishes the proof. $\square$